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24 black birds in a pie

there are 24 blackbirds in a pie!!!!!!!!

Blackbirds Baked in a Pie That's my Guess

24 blackbirds baked in a pie? some times seen as 4 & 20 B B B in a pie.

Do you mean 23 P of C in the H B? 23 Pairs of chromosomes in the human body.

if P(A)>0 then P(B'|A)=1-P(B|A) so P(A intersect B')=P(A)P(B'|A)=P(A)[1-P(B|A)] =P(A)[1-P(B)] =P(A)P(B') the definition of independent events is if P(A intersect B')=P(A)P(B') that is the proof

Sum Rule: P(A) = \sum_{B} P(A,B) Product Rule: P(A , B) = P(A) P(B|A) or P(A, B)=P(B) P(A|B) [P(A|B) means probability of A given that B has occurred] P(A, B) = P(A) P(B) , if A and B are independent events.

P=B×RB=P÷RR=P÷B

There are symbols missing from your question which I cam struggling to guess and re-insert. p(a) = 2/3 p(b ??? a) = 1/2 p(a ∪ b) = 4/5 p(b) = ? Why use the set notation of Union on the third given probability whereas the second probability has something missing but the "sets" are in the other order, and the order wouldn't matter in sets. There are two possibilities: 1) The second probability is: p(b ∩ a) = p(a ∩ b) = 1/2 → p(a) + p(b) = p(a ∪ b) + p(a ∩ b) → p(b) = p(a ∪ b) + p(a ∩ b) - p(a) = 4/5 + 1/2 - 2/3 = 24/30 + 15/30 - 20/30 = 19/30 2) The second and third probabilities are probabilities of "given that", ie: p(b|a) = 1/2 p(a|b) = 4/5 → Use Bayes theorem: p(b)p(a|b) = p(a)p(b|a) → p(b) = (p(a)p(b|a))/p(a|b) = (2/3 × 1/2) / (4/5) = 2/3 × 1/2 × 5/4 = 5/12

P(A|B)= P(A n B) / P(B) P(A n B) = probability of both A and B happening to check for independence you see if P(A|B) = P(B)

If they're disjoint events: P(A and B) = P(A) + P(B) Generally: P(A and B) = P(A) + P(B) - P(A|B)

Blackbirds Baked in a Pie. Should really be 4 and 20!

Let's try this example (best conceived of as a squared 2x2 table with sums to the side). The comma here is an AND logical operator. P(A, B) = 0.1 P(A, non-B) = 0.4 P(non-A, B) = 0.3 P(non-A, non-B) = 0.2 then P(A) and P(B) are obtained by summing on the different sides of the table: P(A) = P(A, B) + P(A, non-B) = 0.1 + 0.4 = 0.5 P(B) = P(A,B) + P(non-A, B) = 0.1 + 0.3 = 0.4 so P(A given B) = P (A, B) / P (B) = 0.1 / 0.4 = 0.25 also written P(A|B) P(B given A) = P (A,B) / P (A) = 0.1 / 0.5 = 0.2 The difference comes from the different negated events added to form the whole P(A) and P(B). If P(A, non-B) = P (B, non-A) then P(A) = P(B) and also P(A|B) = P(B|A).

A LOT. all U.S. aircraft had .50 cals The P-51, P-38, P-47, B-25, B-17, B-29, A-20, B-24, P-39 etc.

This has to do with the union of events. If events A and B are in the set S, then the union of A and B is the set of outcomes in A or B. This means that either event A or event B, or both, can occur. P(A or B) = P(A) + P(B) - P(A and B) **P(A and B) is subtracted, since by taking P(A) + P(B), their intersection, P(A and B), has already been included. In other words, if you did not subtract it, you would be including their intersection twice. Draw a Venn Diagram to visualize. If A and B can only happen separately, i.e., they are independent events and thus P(A and B) = 0, then, P(A or B) = P(A) + P(B) - P(A and B) = P(A) + P(B) - 0 = P(A) + P(B)

If A and B are two events then P(A or B) = P(A) + P(B) - P(A and B)

If A and B are mutually exclusive event then Probability of A or B is P(A)+P(B). If they are not mutually exclusive then it is that minus the probability of the P(A)+P(B) That is to say P( A or B)= P(A)+P(B)- P(A and B). Of course it is clear that if they are mutually exclusive, P(A and B)=0 and we have the first formula.

Given two events, A and B, the conditional probability rule states that P(A and B) = P(A given that B has occurred)*P(B) If A and B are independent, then the occurrence (or not) of B makes no difference to the probability of A happening. So that P(A given that B has occurred) = P(A) and therefore, you get P(A and B) = P(A)*P(B)

P(A'/B)=P(A'nB)/P(B)

Consider events A and B. P(A or B)= P(A) + P(B) - P(A and B) The rule refers to the probability that A can happen, or B can happen, or both can happen together. That is what is stated in the addition rule. Often P(A and B ) is zero, if they are mutually exclusive. In this case the rule just becomes P(A or B)= P(A) + P(B).

Well for one you cant buy Alex Rider the Gemini Project because it doesn't exist what your after is Alex Rider Point Blanc and as for that you can buy it in many places like amazon or got to the libary and get it :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :)

first prove *: if A intersect B is independent, then A intersect B' is independent. (this is on wiki answers) P(A' intersect B') = P(B')P(A'|B') by definition = P(B')[1-P(A|B')] since 1 = P(A) + P(A') = P(B')[1 - P(A)] from the first proof * = P(B')P(A') since 1 = P(A) + P(A') conclude with P(A' intersect B') = P(B')P(A') and is therefore independent by definition. ***note*** i am a student in my first semester of probability so this may be incorrect, but i used the first proof* so i figured i would proof this one to kinda "give back".

The short answer is that if you don't do so, the intersection is double counted. The longer answer follows: Suppose you have two events A and B which overlap. Then, Event A = (only A happens) and (both A and B happen). These two are mutually exclusive so that P(A) = P(A only) + P(both). Similarly, P(B) = P(B only) + P(both). Now, the event that A or B happen is the event that only A happens or only B happens or both happen. That is, P(A or B) = P(A only) + P(both) + P(B only) Adding and subtracting P(both) gives P(A or B) = P(A only) + P(both) + P(B only) + P(both) - P(both) Now, the first two are P(A) and the next two are P(B) So P(A or B) = P(A) + P(B) - P(both).

If B is between P and Q, then: P<B<Q

P(A given B')=[P(A)-P(AnB)]/[1-P(B)].